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Report (any type)/Brief report
22/09/2021
##”700 mL”##
Let’s assume that you’re not familiar with the formula for .
The underlying concept of a dilution is that you can decrease the concentration of a solution by increasing its volume while keeping the number of moles of constant.
Now, a solution’s tells you the number of moles of solute present in one liter of that solution.
##color(blue)(c = n_”solute”/V_”solution”)##
In a dilution, you know that the number of moles of solute must remain constant. This means that you can use the and volume of the initial solution to figure out how many moles it contains
##color(blue)(c = n/V implies n = c xx V)##
##n_(KCl) = 2.4″moles”/color(red)(cancel(color(black)(“L”))) * 500 * 10^(-3)color(red)(cancel(color(black)(“L”))) = “1.2 moles KCl”##
This is exactly how many moles of potassium chloride must be present in the target solution, which means that you have
##color(blue)(c = n/V implies V = n/c)##
##V_”target” = (1.2color(red)(cancel(color(black)(“moles”))))/(1.0color(red)(cancel(color(black)(“moles”)))/”L”) = “1.2 L”##
So, you need the volume of the target solution to be equal to
##V_”target” = “1.2 L” = 1.2 * 10^3″mL” = “1200 mL”##
This means that you must add
##V_”added” = “1200 mL” – “500 mL” = color(green)(“700 mL”)##
to your initial solution to get its molarity down from ##”2.4 M”## to ##”1.0 M”##.
This is exactly how the formula for works
##color(blue)(overbrace(c_1 xx V_1)^(color(brown)(“moles of solute in initial solution”)) = overbrace(c_2 xx V_2)^(color(brown)(“moles of solute in diluted solution”)))##
Here
##c_1##, ##V_1## – the concentration and volume of the initial solution
##c_2##, ##V_2## – the concentration and volume of the target solution
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I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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